2. A child of mass mc is sitting on the edge of a motionless merry-go-round that
ID: 1787655 • Letter: 2
Question
2. A child of mass mc is sitting on the edge of a motionless merry-go-round that has a mass of mp and child can jump at a speed of v Determine the velocity of the child as they land on the ground, and the speed of the merry-go-round if the child jumps off directly away from the center of the ride. (You can ignore vertical motion for this problem.) Determine the velocity of the child as they land on the ground, and the speed of the merry-go-round if the child jumps off tangent to the edge of the ride. (You can ignore vertical motion for this problem.) Determine the velocity of the child as they land on the ground, and the speed of the merry-go-round if the child jumps offat an angle of with respect to the edge of the ride a. b. c.Explanation / Answer
Consider child jumping at an angle theta with the edge, that is, for theta = 0, he jumps along tangent and for theta = 90 , he jumps directly away from center.
Let speed of child be V1 and speed of Merry go round be V2. Let us consider MOI of merry go round MdR2 , that is we consider mass Md as point mass at the edge.
Initial angular momentum of child +merry go round about center is zero.
Angular momentum of child, after jump = Mc V1 R cos theta
Angular momentum of merry-go round in opp. direction = Md V2 R
By angla momentum conservation, final amgular momentum of child + merry go roung is also zero.
Hence Mc V1 R cos theta = Md V2 R .......1
Relative velocity between child and merry go round along the line of velocity of child
Vr= Vj = V1 + V2 cos theta ......2
From equations 1 and 2 we get
V1 = Md Vj / ( Md + Mccos2theta )
V2 = Mccos theta Vj / ( Md + Mccos2theta )
a) Velocity of child as he lands on ground Vg = V1 i + (2gH)1/2j ( i unit vector in horizontal direction and j is init vector in vertical downward direction)
theta = 90 deg
Vg = Vji + (2gH)1/2 j
V2 = 0
b) for theta = 0
Vg = Md Vji / ( Md + Mc) + (2gH)1/2 j
V2 = Mc Vj / ( Md + Mc)
c) for angle theta
Vg = Md Vji/ ( Md + Mccos2theta ) + (2gH)1/2j
V2 = Mccos theta Vj / ( Md + Mccos2theta )
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